Question

Let \(G(V, E)\) be a simple, undirected graph. A vertex cover of \(G\) is a subset \(V' \subseteq V\) such that for every \((u, v) \in E\), \(u \in V'\) or \(v \in V'\). Let the size of the smallest vertex cover in \(G\) be \(k\). Let \(S\) be any vertex cover of size \(k\).
For a vertex \(v \in V\), which of the following constraints will always ensure that \(v \in S\)?

• A. The degree of \(v\) is at least \(k+1\)
• B. The vertex \(v\) is on a path of length \(k+1\)
• C. The vertex \(v\) is on a cycle of length \(k+1\)
• D. The vertex \(v\) is a part of a clique of size \(k\)

Hint:

Use a proof by contradiction on the degree condition, and construct small explicit graphs (a path, a triangle, a complete graph) to test whether the other conditions can fail.

Answer 1

The Correct Option is A

Step 1: Use the vertex-cover / independent-set duality.
A classical fact: \(V' \subseteq V\) is a vertex cover of \(G\) if and only if its complement \(V \setminus V'\) is an independent set of \(G\) (no edge has both endpoints in \(V\setminus V'\), because such an edge would be uncovered by \(V'\)). Consequently, \(S\) is a minimum vertex cover (size \(k\)) if and only if \(V \setminus S\) is a maximum independent set (size \(|V|-k\)).

Step 2: Reformulate the question using this duality.
"\(v \in S\) for every minimum vertex cover \(S\)" is equivalent to "\(v \notin V\setminus S\) for every maximum independent set \(V\setminus S\)", i.e. \(v\) never appears in any maximum independent set of \(G\).

Step 3: Re-examine Option (A) through this lens.
If \(\deg(v) \ge k+1\), could \(v\) belong to some maximum independent set \(I\) (size \(|V|-k\))? If \(v \in I\), then none of \(v\)'s \(\ge k+1\) neighbours can be in \(I\) (independence), so all of them lie in \(V \setminus I\), which has size \(k\). But \(v\) itself is also excluded from \(V\setminus I\), so \(V\setminus I\) would need to contain at least \(k+1\) vertices while having size exactly \(k\) -- a contradiction. So \(v\) cannot be in any maximum independent set, confirming \(v\) is in every minimum vertex cover: Option (A) true, matching Step 2's direct proof.

Step 4: Re-examine Options (B), (C), (D) through this lens using fresh small examples.
For (B): in the path \(v_1v_2v_3v_4\) (\(k=2\)), maximum independent sets of size \(|V|-k=2\) include \(\{v_1,v_3\}\) and \(\{v_1,v_4\}\), both containing \(v_1\) despite \(v_1\) lying on the length-3 path. So \(v_1\) is not excluded from every maximum independent set, hence not forced into every minimum vertex cover: (B) false.
For (C): in the triangle (\(k=2\)), maximum independent sets have size \(|V|-k=1\); \(\{v_1\}\) alone is a valid maximum independent set, so \(v_1\) belongs to a maximum independent set despite lying on the length-3 cycle: (C) false.
For (D): in \(K_{k+1}\), maximum independent sets have size \(|V|-k=1\); any single vertex, including the chosen clique member \(v\), forms a valid maximum independent set \(\{v\}\): (D) false.

Step 5: Conclusion.
The duality-based re-derivation agrees with the direct proof and counterexample method: only Option (A) always forces \(v \in S\). \[ \boxed{\text{Option (A) only}} \]